# Inverse Laplace Transform(s/((s^2)+1)^2

by fysiikka111
Tags: inverse, laplace, transforms or s2
 HW Helper P: 1,583 when you differentiate the integral you take in differentiation under the integral sign: $$\frac{d}{ds}\int_{0}^{\infty}e^{-st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{-st}\sin t\right) dt$$ Differentiating the integrand: $$\frac{d}{ds}e^{-st}\sin t=-te^{-st}\sin t$$ As now we regard s as variable and t is fixed. So the above can be written as: $$(-t\sin t)e^{-st}$$ and so you're taking the Laplace transform of something different now. That something is the answer to your question. Is this making sense now?